What are the solutions to the equation (x)(x-1)=12 ?

The solutions to the equation (x)(x-1)=12 are x=4,x=-3

Find the zeros of (x)(x-1)=12

The zeros of (x)(x-1)=12 are x=4,x=-3

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(x) (x - 1) = 12

x = 4, x = - 3

Solution steps

(x) (x - 1) = 12

Expand

(x) (x - 1) : x^{2} - x

x^{2} - x = 12

Move

12

to the left side

x^{2} - x - 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 1 \cdot ( - 12 )}{2 \cdot 1}

(- 1)^{2} - 4 \cdot 1 \cdot (- 12) = 7

x_{1, 2} = \frac{- ( - 1 ) \pm 7}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 7}{2 \cdot 1}, x_{2} = \frac{- ( - 1 ) - 7}{2 \cdot 1}

x = \frac{- ( - 1 ) + 7}{2 \cdot 1} : 4

x = \frac{- ( - 1 ) - 7}{2 \cdot 1} : - 3

The solutions to the quadratic equation are:

x = 4, x = - 3

3 x - 2 \geq 5 x + 3

4 x - 5 = 2 (x + 5)

72 y + 108 z

2 x + 1 < x - 3

4 x = 0

What are the solutions to the equation (x)(x-1)=12 ?
The solutions to the equation (x)(x-1)=12 are x=4,x=-3
Find the zeros of (x)(x-1)=12
The zeros of (x)(x-1)=12 are x=4,x=-3